Problem: A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters + and 1 into this sequence. For example, sequences (())(), () and (()(())) are regular, while )(, (() and (()))( are not. Let's call a regular bracket sequence "RBS".

You are given a sequence s of n characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.

You have to replace every character ? with either ) or ( (different characters ? can be replaced with different brackets). You cannot reorder the characters, remove them, insert other characters, and each ? must be replaced.

Determine if it is possible to obtain an RBS after these replacements.

Input

The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.

Each test case consists of one line containing s (2 ≤ |s| ≤ 100) — a sequence of characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.

Output

For each test case, print YES if it is possible to obtain a regular bracket sequence, or NO otherwise}. 

You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).

Example

Input


5
()
(?)
(??)
??()
)?(?


Output


YES
NO
YES
YES
NO

Note

In the first test case, the sequence is already an RBS.

In the third test case, you can obtain an RBS as follows: ()() or (()).

In the fourth test case, you can obtain an RBS as follows: ()().

Solution (in PYTHON3):

Solution:
import math
from sys import *
#input=stdin.readline

def solve():
    #n=int(input())
    
    s=input()
    #n,k=map(int,input().split())
    ##m=int(input())
    #l=list(map(int,input().split()))
    #l1=list(map(int,input().split()))
    #zero=s.count('0')
    n=len(s)
    if(s[n-1]=='(' or s[0]==')'):
        print("NO")
    else:
        c1,c2,c3=0,0,0
        for i in s:
            if(i=='('):
                c1+=1
            elif(i==')'):
                c2+=1
            elif(i=='?'):
                c3+=1
        if(abs(c1-c2)==0 and c3%2!=0):
            print("NO")
        elif(abs(c1-c2)!=0 and abs(c1-c2)!=c3):
            print("NO")
        else:
            print("YES")
    

#t=1    
t=int(input())
while(t>0):
    t-=1
    solve()